Question: The radius of a sphere is increasing at a rate of $0.5$ centimeters per minute. At a certain instant, the radius is $17$ centimeters. What is the rate of change of the volume of the sphere at that instant (in cubic centimeters per minute)? Choose 1 answer: Choose 1 answer: (Choice A) A $3275\pi$ (Choice B) B $\dfrac16\pi$ (Choice C) C $289\pi$ (Choice D) D $578\pi$ The volume of a sphere with radius $r$ is $\dfrac{4}{3}\pi r^3$.
Answer: Setting up the math Let... $r(t)$ denote the sphere's radius at time $t$, and $V(t)$ denote the sphere's volume at time $t$. We are given that $r'(t)=0.5$ and that $r(t_0)=17$ for a specific time $t_0$. We want to find $V'(t_0)$. Relating the measures $V(t)$ and $r(t)$ relate to each other through the formula for the volume of a sphere: $V(t)=\dfrac43\pi[r(t)]^3$ We can differentiate both sides to find an expression for $V'(t)$ : $V'(t)=4\pi[r(t)]^2r'(t)$ Using the information to solve Let's plug ${r(t_0)}={17}$ and ${r'(t_0)}={0.5}$ into the expression for $V'(t_0)$ : $\begin{aligned} V'(t_0)&=4\pi[{r(t_0)}]^2{r'(t_0)} \\\\ &=4\pi({17})^2({0.5}) \\\\ &=578\pi \end{aligned}$ In conclusion, the rate of change of the volume of the sphere at that instant is $578\pi$ cubic centimeters per minute. Since the rate of change is positive, we know that the volume is increasing.